Báo cáo hóa học: "WEAK AND STRONG CONVERGENCE OF FINITE FAMILY WITH ERRORS OF NONEXPANSIVE NONSELF-MAPPINGS" docx

UNGCHITTRAKOOL Received 27 September 2005; Revised 5 May 2006; Accepted 8 May 2006 We are concerned with the study of a multistep iterative scheme with errors involving a finite family o

WITH ERRORS OF NONEXPANSIVE NONSELF-MAPPINGS

S PLUBTIENG AND K UNGCHITTRAKOOL

Received 27 September 2005; Revised 5 May 2006; Accepted 8 May 2006

We are concerned with the study of a multistep iterative scheme with errors involving

a finite family of nonexpansive nonself-mappings We approximate the common fixed points of a finite family of nonexpansive nonself-mappings by weak and strong conver-gence of the scheme in a uniformly convex Banach space Our results extend and improve some recent results, Shahzad (2005) and many others

Copyright © 2006 S Plubtieng and K Ungchittrakool This is an open access article dis-tributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is prop-erly cited

1 Introduction

LetK be a subset of a real normed linear space E and let T be a self-mapping on K T is

said to be nonexpansive provided Tx − T y  x − y for allx, y ∈ K.

Fixed-point iteration process for nonexpansive mappings in Banach spaces includ-ing Mann and Ishikawa iteration processes has been studied extensively by many au-thors to solve the nonlinear operator equations in Hilbert spaces and Banach spaces; see [3,7,10,11,15,16] Tan and Xu [15] introduced and studied a modified Ishikawa process to approximate fixed points of nonexpansive mappings defined on nonempty closed convex bounded subsets of a uniformly convex Banach spaceE Five years later,

Xu [18] introduced iterative schemes known as Mann iterative scheme with errors and Ishikawa iterative scheme with errors Takahashi and Tamura [14] introduced and stud-ied a generalization of Ishikawa iterative schemes for a pair of nonexpansive mappings

in Banach spaces Recently, Khan and Fukhar-ud-din [6] extended their scheme to the modified Ishikawa iterative schemes with errors for two mappings and gave weak and strong convergence theorems On the other hand, iterative techniques for approximat-ing fixed points of nonexpansive nonself-mappapproximat-ings have been studied by various au-thors; see [4,8,13,19] Shahzad [12] introduced and studied an iteration scheme for

Hindawi Publishing Corporation

Fixed Point Theory and Applications

Volume 2006, Article ID 81493, Pages 1 12

DOI 10.1155/FPTA/2006/81493

approximating a fixed point of nonexpansive nonself-mappings (when such a fixed point exists) and gave some strong and weak convergence theorems for such mappings Inspired and motivated by these facts, we introduce and study a multistep iterative scheme with errors for a finite family of nonexpansive nonself-mappings Our schemes can be viewed as an extension for two-step iterative schemes of Shahzad [12] The scheme

is defined as follows

LetK be a nonempty closed convex subset of a uniformly convex Banach space E,

which is also a nonexpansive retract ofE And let T1,T2, , T N:K → E be nonexpansive

mappings, the following iteration scheme is studied:

x1

n = P

α1

n T1x n+β1

n x n+γ1

n u1

n

 ,

x2

n = P

α2

n T2x1

n+β2

n x n+γ2

n u2

n

 ,

x n+1 = x N

n = P

α N

n T N x N

n −1 +β N

n x n+γ N

n u N n



(1.1)

withx1∈ K, n1, whereP is a nonexpansive retraction with respect to K and { α1

n },

{ α2

n }, , { α N

n },{ β1

n },{ β2

n }, , { β N

n },{ γ1

n },{ γ2

n }, , { γ N

n }are sequences in [0, 1] withα i

n+

β i

n+γ i

n =1 for alli =1, 2, , N, and { u1

n },{ u2

n }, , { u N

n }are bounded sequences inK.

ForN =2,T1= T2≡ T, β n = α1

n,α n = α2

n, andγ1

n = γ2

n ≡0, then (1.1) reduces to the scheme for a mapping defined by Shahzad [12]:

x1= x ∈ K,

x n+1 = P

1− α n

x n+α n TP

1− β n

x n+β n Tx n

where{ α n },{ β n }are sequences in [0, 1]

For N =2,T1,T2:K → K, T1= T, T2= S, and y n = x1

n, then (1.1) reduces to the scheme with errors for two mappings defined by

x1= x ∈ K,

y n = α1

n Tx n+β1

n x n+γ1

n u1

n,

x n+1 = x2

n = α2

n Sy n+β2

n x n+γ2

n u2

n,

(1.3)

where{ α1

n },{ α2

n },{ β1

n },{ β2

n },{ γ1

n },{ γ2

n }are sequences in [0, 1] withα1

n+β1

n+γ1

n =1= α2

n+

β2

n+γ2

nand{ u1

n },{ u2

n }are bounded sequences inK.

It is our purpose in this paper to establish several weak and strong convergence theorems of the multistep iterative scheme with errors for a finite family of nonexpansive nonself-mappings More precisely, we prove weak convergence of these implicit iteration

processes in a uniformly convex Banach space which has the Kadec-Klee property The results presented in this paper extend and improve the corresponding ones announced

by Shahzad [12], and many others

2 Preliminaries

In this section, we recall the well-known concepts and results

LetE be a real Banach space A subset K of E is said to be a retract of E if there exists

a continuous mapP : E → K such that Px = x for all x ∈ K A map P : E → E is said to be

a retraction if P2= P It follows that if a map P is a retraction, then P y = y for all y in

the range ofP A mapping T : K → E is called demiclosed with respect to y ∈ E if for each

sequence{ x n }inK and each x ∈ E, x n  x and Tx n → y imply that x ∈ K and Tx = y A

Banach spaceE is said to satisfy Opial’s condition [9] if for any sequence{ x n }inE, x n  x

implies that

lim sup

n →∞

x n − x< lim sup

n →∞

for ally ∈ E with x = y A Banach space E is said to have the Kadec-Klee property if for

every sequence{ x n }inE, x n  x and  x n  →  x together imply x n − x  →0 A family

{ T i: =1, 2, , N }ofN nonself-mappings of K (i.e., T i:K → E) with F = N

i =1F(T i)=∅

is said to satisfy condition (B) on K if there is a nondecreasing function f : [0, ∞)→[0,∞) with f (0) =0 andf (r) > 0 for all r ∈(0,∞) such that for allx ∈ K,

max

1iN x − T i xf

d(x, F)

The family{ T i: =1, 2, , N } is said to satisfy condition (A N) if (2.2) is replaced by

1/NN

i =1 x − T i x f (d(x, F)) for all x ∈ K Note that condition (B) reduces to

condi-tion (A N) when x − T1x  =  x − T2x  = ··· =  x − T N x 

A mappingT : K → E is called semicompact if any sequence { x n }inK satisfying  x n −

Tx n  →0 asn → ∞has a convergent subsequence

Lemma 2.1 (Tan and Xu [15]) Let { s n } , { t n } be two nonnegative sequences satisfying

If∞

n =1t n < ∞ , then lim n →∞ s n exists Moreover, if there exists a subsequence { s n j } of { s n }

such that s n j → 0 as j → ∞ , then s n → 0 as n → ∞

Lemma 2.2 (Xu [17]) Let p > 1 and R > 0 be two fixed numbers and E a Banach space Then E is uniformly convex if and only if there exists a continuous, strictly increasing, and convex function g : [0, ∞)→[0,∞ ) with g(0) = 0 such that  λx + (1 − λ)y  pλ  x  p+ (1− λ)  y  p − W p(λ)g(  x − y  ) for all x, y ∈ B R(0)= { x ∈ E :  x R } , and λ ∈ [0, 1], where W p(λ) = λ(1 − λ) p+λ p(1− λ).

Lemma 2.3 (Kaczor [5]) Let E be a real reflexive Banach space such that its dual E ∗ has the Kadec-Klee property Let { x n } be a bounded sequence in E and x ∗,y ∗ ∈ ω w(x n ); here ω w(x n)

denote the set of all weak subsequential limits of { x n } Suppose lim n →∞  tx n+(1− t)x ∗ − y ∗ 

exists for all t ∈ [0, 1] Then x ∗ = y ∗

Lemma 2.4 (Browder [1]) Let E be a uniformly convex Banach space, K a nonempty closed convex subset of E, and T : K → E a nonexpansive mapping Then I − T is demiclosed at zero.

3 Main results

In this section, we prove weak and strong convergence theorems of the iterative scheme given in (1.1) for a finite family of nonexpansive mappings in a Banach space In order to prove our main results, the following lemmas are needed

Lemma 3.1 Let E be a uniformly convex Banach space and K a nonempty closed con-vex subset of E which is also a nonexpansive retract of E Let T1,T2, , T N:K → E be nonexpansive mappings Let { x n } be the sequence defined by ( 1.1 ) with ∞

n =1γ i

n < ∞ for each i =1, 2, , N If N

i =1F(T i)=∅, then lim n →∞  x n − x ∗  exists for all x ∗ ∈ N

i =1F(T i ) Proof For each n1, we note that

x1

n − x ∗α1

nT1x n − x ∗+β1

nx n − x ∗+γ1

nu1

n − x ∗

α1nx n − x ∗+β1

nx n − x ∗+γ1

nu1

n − x ∗

x n − x ∗+d0

n,

(3.1)

whered0

n = γ1

n  u1

n − x ∗  Since∞

n =1γ1

n < ∞,∞

n =1d0

n < ∞ Next, we note that

x2

n − x ∗α2

nT2x1

n − x ∗+β2

nx n − x ∗+γ2

nu2

n − x ∗

α2nx1

n − x ∗+β2

nx n − x ∗+γ2

nu2

n − x ∗

=α2

n+β2

nx n − x ∗+α2

n d0

n+γ2

nu2

n − x ∗

x n − x ∗+d1

n,

(3.2)

whered1

n = α2

n d0

n+γ2

n  u2

n − x ∗  Since∞

n =1d0

n < ∞and∞

n =1γ2

n < ∞,∞

n =1d1

n < ∞ Simi-larly, we have

x3

n − x ∗α3

nx2

n − x ∗+β3

nx n − x ∗+γ3

nu3

n − x ∗

α3nx n − x ∗+d1

n

 +β3nx n − x ∗+γ3

nu3

n − x ∗

x n − x ∗+α3

n d n1+γ3nu3

n − x ∗  =  x n − x ∗+d2

n,

(3.3)

whered2

n = α3

n d1

n+γ3

n  u3

n − x ∗ , so∞

n =1d2

n < ∞

By continuing the above method, there exists a nonnegative real sequence{ d i −1

n }such that∞

n =1d i −1

n < ∞and

x i

n − x ∗x n − x ∗+d i −1

n , ∀ n1,∀ i =1, 2, , N. (3.4) Thus x n+1 − x ∗  =  x N

n − x ∗  x n − x ∗ +d N −1

n for alln ∈ N Hence, byLemma 2.1, limn →∞  x n − x ∗ exists This completes the proof 

Lemma 3.2 Let E be a uniformly convex Banach space and K a nonempty closed convex subset of E which is also a nonexpansive retract of E Let T1,T2, , T N:K → E be nonex-pansive mappings Let { x n } be the sequence defined by ( 1.1 ) with∞

n =1γ i

n < ∞ and { α i

n } ⊆

[ε, 1 − ε] for all i =1, 2, , N, for some ε ∈ (0, 1) If N

i =1F(T i)=∅, then lim n →∞  x n −

T i x n  = 0 for all i =1, 2, , N.

Proof Let x ∗ ∈ N

i =1F(T i) Then, by Lemma 3.1, limn →∞  x n − x ∗ exists Let limn →∞ 

x n − x ∗  = r If r =0, then by the continuity of eachT i the conclusion follows Sup-pose thatr > 0 Firstly, we are now to show that lim n →∞  T N x n − x n  =0 Since{ x n }and

{ u i

n }are bounded for alli =1, 2, , N, there exists R > 0 such that x n − x ∗+γ i

n(u i

n − x n),

T i x i −1

n − x ∗+γ i

n(u i

n − x n)∈ B R(0) for alln1 and for alli =1, 2, , N UsingLemma 2.2,

we have

x N

n − x ∗ 2

α N

n T N x N −1

n +β N

n x n+γ N

n u N

n − x ∗ 2

=α N

n



T N x N −1

n − x ∗+γ N

n



u N

n − x n

+

1− α N n



x n − x ∗+γ N

n



u N

n − x n 2

α N

nT N x N −1

n − x ∗+γ N

n



u N

n − x n 2

+

1− α N

nx n − x ∗+γ N

n



u N

n − x n 2

− W2 

α N n

gT N x N −1

n − x n

α N nx N −1

n − x ∗+γ N

nu N

n − x n 2

+

1− α N nx n − x ∗+γ N

nu N

n − x n 2

− W2



α N n



gT N x N −1

n − x n

α N nx n − x ∗+d N −2

n +γ N nu N

n − x n 2 +

1− α N nx n − x ∗+d N −2

n +γ n Nu N

n − x n 2

− W2



α N n



gT N x N −1

n − x n

=x n − x ∗+λ N −2

n

 2

− W2



α N n



gT N x N −1

n − x n,

(3.5) whereλ N −2

n := d N −2

n +γ N

n  u N

n − x ∗  Observe thatε3W2(α N

n) now (3.5) implies that

ε3g(  T N x N −1

n − x n ) x n − x ∗ 2−  x n+1 − x ∗ 2+ρ N −2

n , where ρ N −2

n :=2 N −2

n  x n −

x ∗ 2+ (λ N −2

n )2 Since∞

n =1d N −2

n < ∞and∞

n =1γ N −2

n < ∞, we get∞

n =1ρ N −2

n < ∞ This implies that limn →∞ g(  T N x N −1

n − x n )=0 Sinceg is strictly increasing and continuous

at 0, it follows that limn →∞  T N x N −1

n − x n  =0 Note that

x n − x ∗x n − T N x N −1

n +T N x N −1

n − x ∗

x n − T N x N −1

n +x N −1

for alln1 Thusr =limn →∞  x n − x ∗ lim infn →∞  x N −1

n − x ∗ lim supn →∞  x N −1

n −

x ∗ r and therefore lim n →∞  x N −1

n − x ∗  = r Using the same argument in the proof

above, we have

x N −1

n − x ∗ 2

α N n −1x N −2

n − x ∗+γ N −1

n u N −1

n − x ∗ 2 +

1− α N −1

n x n − x ∗+γ N −1

n u N −1

n − x ∗ 2

− W2 

α N −1

n



gT N −1x N −2

n − x n

α N −1

n x n − x ∗+d N −3

n +γ N −1

n u N −1

n − x ∗ 2 +

1− α N −1

n x n − x ∗+d N −3

n +γ N −1

n u N −1

n − x ∗ 2

− W2



α N −1

n



gT N −1x N −2

n − x n

x n − x ∗ 2

+ρ N n −3− W2



α N n −1

gT N −1x N −2

n − x n.

(3.7)

This implies thatε3g(  T N −1x N −2

n − x n ) x n − x ∗ 2−  x N −1

n − x ∗ 2+ρ N −3

n and there-fore limn →∞  T N −1x N −2

n − x n  =0 Thus, we have

x n − T N x nx n − T N x N −1

n +T N x N −1

n − T N x n

x n − T N x N −1

n +x N −1

n − x n

=x n − T N x N −1

n +P

α N −1

n T N −1x N −2

n +β N −1

n x n+γ N −1

n u N −1

n



− Px n

x n − T N x N −1

n +α N −1

n x n − T N −1x N −2

n +γ N −1

n u N −1

n − x n.

(3.8)

Since limn →∞  x n − T N x N −1

n  =0, limn →∞  x n − T N −1x N −2

n  =0, and∞

n =1γ N −1

n < ∞, it follows that limn →∞  x n − T N x n  =0 Similarly, by using the same argument as in the proof above, we have limn →∞  x n − T N −2x N −3

n  =limn →∞  x n − T N −3x N −4

n  =, , =

limn →∞  x n − T2x1

n =0 This implies that limn →∞  x n − T N −1x n =limn →∞  x n − T N −2x n  =

, , =limn →∞  x n − T3x n  =0 It remains to show that

limn →∞x n − T1x n  =0, limn →∞x n − T2x n  =0. (3.9)

Note that

x1

n − x ∗ 2

α1

nx n − x ∗+γ1

nu1

n − x ∗ 2 +

1− α1

nx n − x ∗+γ1

nu1

n − x ∗ 2

− W2



α1

n



gT1x n − x n

=x n − x ∗+γ1

nu1

n − x ∗ 2

− W2



α1n

gT1x n − x n.

(3.10)

Thus, we haveε3g(  T1x n − x n )( x n − x ∗ +γ1

n  u1

n − x ∗ )2−  x1

n − x ∗ 2 and there-fore limn →∞  T1x n − x n =0 Since x n − T2x n  x n − T2x1

n +α1

n  T1x n − x n +γ1

n  u1

n −

x n , it implies that limn →∞  T2x n − x n  =0 Therefore limn →∞  T i x n − x n  =0 for all

Theorem 3.3 Let E be a uniformly convex Banach space and let K be a nonempty closed convex subset of E which is also a nonexpansive retract of E Let T1,T2, , T N:K → E be nonexpansive mappings which are satisfying condition (B) Let { x n } be the sequence defined

by ( 1.1 ) with∞

n =1γ i

n < ∞ and { α i

n } ⊆[ε, 1 − ε] for all i =1, 2, , N for some ε ∈ (0, 1) If

F : = N

i =1F(T i)=∅, then { x n } converges strongly to a common fixed point in F.

Proof ByLemma 3.2, limn →∞  T i x n − x n  =0 for alli =1, 2, , N Now by condition (B),

f (d(x n,F))M n:=max1iN { T i x n − x n }for alln ∈ N Hence limn →∞ f (d(x n,F)) =

0 Since f is a nondecreasing function and f (0) =0, therefore limn →∞ d(x n,F) =0 Now we can choose a subsequence{ x n j }of{ x n }and a sequence{ y j } ∈ F such that

 x n j − y j  < 2 − j By the following method of the proof of Tan and Xu [15], we get that

{ y j }is a Cauchy sequence inF and so it converges Let y j → y Since F is closed, therefore

y ∈ F and then x n j → y ByLemma 3.1, limn →∞  x n − x ∗ exists for allx ∗ ∈ F, x n → y ∈

ForN =2,T1= T2≡ T, β n = α1

n,α n = α2

n, andγ1

n = γ2

n ≡0 inTheorem 3.3, we obtain the following results

Corollary 3.4 (see [12, Theorem 3.6]) Let E be a real uniformly convex Banach space and K a nonempty closed convex subset of E which is also a nonexpansive retract of E Let

T : K → E be a nonexpansive mapping with F(T) =∅ Let { α n } and { β n } be sequences

in [ε, 1 − ε] for some ε ∈ (0, 1) From arbitrary x1∈ K, define the sequence { x n } by the recursion ( 1.2 ) Suppose T satisfies condition (A1) Then { x n } converges strongly to some fixed point of T.

WhenN =2,S = T1,T = T2:C → C, and y n = x1

ninTheorem 3.3, we obtain strong convergence theorem as follows

Corollary 3.5 Let E be a uniformly convex Banach space and let C be a nonempty closed convex subset of E which is also a nonexpansive retract of E Let S, T be nonexpan-sive mappings of C into itself satisfying condition (A2), and let { x n } be sequence defined

by ( 1.3 ) with∞

n =1γ1

n < ∞ ,∞

n =1γ2

n < ∞ and 0 < δα1

n , α2

n1− δ < 1 for all n ∈ N If

F : = F(S) ∩ F(T) =∅, then { x n } converges strongly to a common fixed point of S and T.

Theorem 3.6 Let E be a uniformly convex Banach space and let K be a nonempty closed convex subset of E which is also a nonexpansive retract of E Let T1,T2, , T N:K → E be nonexpansive mappings Suppose that one of the mappings in { T i: =1, 2, , N } is semi-compact Let { x n } be the sequence defined by ( 1.1 ) with∞

n =1γ i

n < ∞ and { α i

n } ⊆[ε, 1 − ε] for all i =1, 2, , N for some ε ∈ (0, 1) If F : = N

i =1F(T i)=∅, then { x n } converges strongly

to a common fixed point in F.

Proof Suppose that T i0is semicompact for some i0=1, 2, , N ByLemma 3.1, we have limn →∞  x n − T i0x n  =0 So there exists a subsequence{ x n j } of { x n } such that x n j →

x ∗ ∈ K as j → ∞ NowLemma 3.2guarantees that limj →∞  x n j − T l x n j  =0 for alll =

1, 2, , N and so  x ∗ − T l x ∗  =0 for all l =1, 2, , N This implies that x ∗ ∈ F By

Lemma 3.1, limn →∞  x n − x ∗ exists and then limn →∞  x n − x ∗  =limj →∞  x n j − x ∗  =0

Theorem 3.7 Let E be a uniformly convex Banach space satisfying the Opial’s condition and K a nonempty closed convex subset of E which is also a nonexpansive retract of E Let

T1,T2, , T N:K → E be nonexpansive mappings and let { x n } be a sequence defined by ( 1.1 ) with∞

n =1γ i

n < ∞ and { α i

n } ⊆[ε, 1 − ε] for all i =1, 2, , N for some ε ∈ (0, 1) If F : =

N

i =1F(T i)=∅, then { x n } converges weakly to a common fixed point in F.

Proof Let x ∗ ∈ F Then as proved inLemma 3.1, limx →∞  x n − x ∗ exists Now we prove that { x n }has a unique weak subsequential limit in F To prove this, let x n i  z1 and

x n j  z2for some subsequences{ x n i },{ x n j }of{ x n } ByLemma 3.2,

lim

i →∞x n

i − T k x n i =0=lim

j →∞x n

j − T k x n j (3.11)

for allk =1, 2, , N and byLemma 2.4 insures thatI − T k are demiclosed at zero for all k =1, 2, , N Therefore we obtain T k z1= z1 and T k z2= z2 for all k =1, 2, , N.

Thenz1,z2∈ F Next, we prove the uniqueness Suppose that z1= z2, then by the Opial’s condition limn →∞  x n − z1 =limi →∞  x n i − z1 < lim i →∞  x n i − z2 =limj →∞  x n j − z2 <

limj →∞  x n j − z1 =limn →∞  x n − z1 This is a contradiction Hence { x n } converges

Lemma 3.8 Let E be a real uniformly convex Banach space and K a nonempty closed convex subset of E which is also a nonexpansive retract of E Let T1,T2, , T N:K → E be nonexpan-sive mappings From arbitrary x1∈ K, define the sequence { x n } by the recursion ( 1.1 ) with for each i =1, 2, , N,∞

n =1γ i

n < ∞ If F : = N

i =1F(T i)=∅, then for all u, v ∈ F, the limit

lim

n →∞tx n+ (1− t)u − v (3.12)

exists for all t ∈ [0, 1].

Proof ByLemma 3.1, we have limn →∞  x n − x ∗ exists for allx ∗ ∈ F This implies that

{ x n }is bounded Observe that there existsR > 0 such that { x n } ⊂ C : = B R(0)∩ K, and

henceC is a nonempty closed convex bounded subset of E Let a n(t) : = tx n+(1− t)u − v  Then limn →∞ a n(0)=  u − v , and fromLemma 3.1, limn →∞ a n(1)=limn →∞  x n − v  ex-ists Without loss of generality, we may assume that limn →∞  x n − u  = r > 0 and

t ∈(0, 1) For anyn1 and for alli =1, 2, , N, we define A i

n:C → C by

A1n:= P

α1n T1+β1n I + γ1n u1n

,

A2

n:= P

α2

n T2A1

n+β2

n I + γ2

n u2

n

 ,

A N n := P

α N n T N A N n −1+β N n I + γ N n u N n

.

(3.13)

Thus, for all x, y ∈ K, we have  A i

n x − A i

n y α i

n  A i −1

n x − A i −1

n y +β i

n  x − y for all

i =2, , N, and  A1

n x − A1

n y α1

n  x − y +β1

n  x − y  This implies that

A N

n x − A N

SetS n,m:= A N n+m −1A N n+m −2··· A N

n,m1, andb n,m:=  S n,m(tx n+ (1− t)u) −(tS n,m x n+ (1− t)S n,m u)  It easy to see thatA N

n x n = x n+1,S n,m x n = x n+m, and S n,m x − S n,m y  x − y 

We show first that, for anyx ∗ ∈ F,  S n,m x ∗ − x ∗  →0 uniformly for allm1 asn →

∞ Indeed, for anyx ∗ ∈ F, we have

A i

n x ∗ − x ∗α i

nA i −1

n x ∗ − x ∗+γ i

nu i

for alli =2, , N, and  A1

n x ∗ − x ∗ γ1

n  u1

n − x ∗  Therefore

A N

n x ∗ − x ∗σ2

n γ1

nu1

n − x ∗+σ3

n γ2

nu2

n − x ∗+···+σ N

n γ N −1

n u N −1

n − x ∗

+γ N

nu N

n − x ∗M N

i =1

γ i

n,

(3.16)

for all n1, where M =max{supn1{ u1

n − x ∗ }, , sup n1{ u N

n − x ∗ }} and σ k

n =

N

i = k α i

n Hence

S n,m x ∗ − x ∗A N

n+m −1A N n+m −2··· A N

n x ∗ − A N

n+m −1A N n+m −2··· A N

n+1 x ∗

+A N n+m −1A N n+m −2··· A N n+1 x ∗ − A N n+m −1A N n+m −2··· A N n+2 x ∗

+A N n+m −1x ∗ − x ∗

A N

n x ∗ − x ∗+A N

n+1 x ∗ − x ∗+···+A N

n+m −1x ∗ − x ∗

M

N

i =1



γ i

n+γ i n+1+···+γ i

n+m −1



M

N

i =1

∞

k = n

γ i k:= δ n x ∗

(3.17)

n =1γ i

n < ∞, for alli =1, 2, , N, we have δ x ∗

n →0 asn →∞and hence S n,m x ∗ − x ∗ →0

asn → ∞ Observe that

a n+m(t) =tS n,m x n+ (1− t)u − v

tS n,m x n+ (1− t)u − S n,m

tx n+ (1− t)u

+S n,m

tx n+ (1− t)u

− v

=tS n,m x n+ (1− t)S n,m u − S n,m

tx n+ (1− t)u

+ (1− t)

u − S n,m u

+S n,m

tx n+ (1− t)u

− v

b n,m+S n,m

tx n+ (1− t)u

− v+ (1− t)u − S n,m u

b n,m+S n,m

tx n+ (1− t)u

− S n,m v+S n,m v − v

+ (1− t)u − S n,m u

b n,m+a n(t) +S n,m v − v+ (1− t)u − S n,m u

b n,m+a n(t) + δ n v+ (1− t)δ n u

(3.18)

By using [2, Theorem 2.3], we have

b n,mϕ −1x n − u − S n,m x n − S n,m u

= ϕ −1x n − u − x n+m − u + u − S n,m u

ϕ −1x n − u − x n+m − u − S n,m u − u, (3.19) and so the sequence{ b n,m }converges uniformly to 0 asn → ∞for allm1 Thus, fixing

n and letting m → ∞in (3.19), we have

lim sup

m →∞ a n+m(t)ϕ −1 

x n − u −lim

m →∞x m − u − δ u

+a n(t) + δ v

n+ (1− t)δ u

(3.20) and again lettingn → ∞,

lim sup

n →∞ a n(t)ϕ −1(0) + lim inf

n →∞ a n(t) + 0 + 0 =lim inf

n →∞ a n(t). (3.21)

Theorem 3.9 Let E be a real uniformly convex Banach space such that its dual E ∗ has the Kaded-Klee property and K a nonempty closed convex subset of E which is also a nonexpan-sive retract of E Let T1,T2, , T N:K → E be nonexpansive mappings with F : = N

i =1F(T i)=

∅ From arbitrary x1∈ K, define the sequence { x n } by the recursion ( 1.1 ) with for each

i =1, 2, , N, ∞

n =1γ i

n < ∞ and α i

n ∈[ε, 1 − ε] for some ε ∈ (0, 1) Then { x n } converges weakly to some fixed point of T.

Proof. Lemma 3.1guarantees that { x n } is bounded SinceE is reflexive, there exists a

subsequence{ x n }of{ x n }converging weakly to somex ∗ ∈ K ByLemma 3.2, we have

3 Main results

In this section, we prove weak and strong convergence. .. convex Banach space and K a nonempty closed convex subset of E which is also a nonexpansive retract of E Let

T : K → E be a nonexpansive mapping with F(T) =∅... strong convergence theorem as follows

Corollary 3.5 Let E be a uniformly convex Banach space and let C be a nonempty closed convex subset of E which is also a nonexpansive retract of